TITLE: Do $\mathbb{HP}^2\#\overline{\mathbb{HP}^2}$ and $\mathbb{OP}^2\#\overline{\mathbb{OP}^2}$ arise as sphere bundles over spheres?
QUESTION [11 upvotes]: Recall that $\mathbb{RP}^2\#\mathbb{RP}^2$ is the Klein bottle and can be seen as a non-trivial $S^1$-bundle over $S^1$. In particular, it is the total space of the sphere bundle of $\gamma\oplus\varepsilon^1$ over $S^1$ where $\gamma$ is the Möbius line bundle and $\varepsilon^1$ is the trivial line bundle; note, $\gamma\oplus \varepsilon^1$ is the unique non-trivial rank two vector bundle on $S^1$ (because $\pi_0(O(2)) = \mathbb{Z}_2$) and it is not orientable (i.e. $w_1(\gamma\oplus\varepsilon^1) \neq 0$) as is $K$.
Also $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ can be seen as a non-trivial $S^2$-bundle over $S^2$. One way to do this is to view it as the first Hirzebruch surface $\mathbb{P}(\mathcal{O}(-1)\oplus\mathcal{O})$; here $\mathcal{O}(-1)$ and $\mathcal{O}$ are holomorphic line bundles over $\mathbb{CP}^1$. Alternatively, it is the total space of the sphere bundle of $\mathcal{O}(-1)\oplus\varepsilon^1$ over $S^2$. Note that $\mathcal{O}(-1)\oplus\varepsilon^1$ is the unique non-trivial rank three vector bundle on $S^2$ (because $\pi_1(SO(3)) = \mathbb{Z}_2$) and it is not spin (i.e. $w_2(\mathcal{O}(1)\oplus\varepsilon^1)) \neq 0$) as is $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. 
This naturally leads one to ask:

Can $\mathbb{HP}^2\#\overline{\mathbb{HP}^2}$ and $\mathbb{OP}^2\#\overline{\mathbb{OP}^2}$ be realised as (non-trivial) sphere bundles over $S^4$ and $S^8$ respectively?

Not all sphere bundles in these dimensions necessarily arise as sphere bundles of vector bundles. However, given the constructions above, I would expect them to in this case.
Unlike in the real and complex cases, there is not a unique non-trivial vector bundle to take the sphere bundle of because $\pi_3(SO(5)) = \mathbb{Z}$ and $\pi_7(SO(9)) = \mathbb{Z}$. All of these homotopy groups are in the stable range, so it is no coincidence that there we have two $\mathbb{Z}_2$'s and two $\mathbb{Z}$'s appearing. As $w_4(\mathbb{HP}^2\#\overline{\mathbb{HP}^2}) \neq 0$ and $w_8(\mathbb{OP}^2\#\overline{\mathbb{OP}^2}) \neq 0$, the desired vector bundles should have non-zero $w_4$ and $w_8$ respectively (in direct analogy with the cases above).
Note that both $\gamma$ and $\mathcal{O}(-1)$ can be viewed as the tautological (real/complex) line bundles over $\mathbb{RP}^1$ and $\mathbb{CP}^1$ respectively. This suggests that we should try the direct sum of the tautological (quaternionic/octionic) line bundles over $\mathbb{HP}^2$ and $\mathbb{OP}^2$ with $\varepsilon^1$. This is perfectly reasonable in the quaternionic case, but $\mathbb{OP}^2$ does not arise as the projectivisation of $\mathbb{O}^3$, so there is no such bundle; maybe there is a natural replacelement though (probably associated to one of the generators of $\pi_7(SO(8)) \cong \mathbb{Z}\oplus\mathbb{Z}$).

REPLY [5 votes]: Another way of describing the content of the previous answers is as follows.  Suppose one starts with one of the Hopf bundles $p:S^{2n-1}\to S^n$ for $n=2,4,8$ coming from $\mathbb C$, $\mathbb H$, or $\mathbb O$, with fibers $S^{n-1}$. The mapping cone of $p$ is then the associated projective plane $\mathbb {CP}^2$, $\mathbb {HP}^2$, or $\mathbb {OP}^2$. The mapping cylinder is obtained from the mapping cone by deleting an open cone on the domain $S^{2n-1}$ of $p$ so this amounts to deleting the interior of a ball $B^{2n}$. The mapping cylinder of a fiber bundle with fibers $S^{n-1}$ is a fiber bundle with fibers $B^n$ over the same base space, $S^n$ in this case. The double of this mapping cylinder is then a bundle over the same base with fibers $S^n$, the double of $B^n$.  On the other hand, the double is also the connected sum of two copies of the projective plane since we are taking taking two copies of the projective plane, deleting the interior of a ball $B^{2n}$ from each, then identifying the two resulting boundary spheres $S^{2n-1}$.